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04 Aug 2021, 02:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
73% (01:22) correct
27%
(01:26)
wrong
based on 74
sessions
History
Date
Time
Result
Not Attempted Yet
The number obtained by interchanging the two digits of a two-digit number is lesser than the original number by 54. If the sum of the two digits of the number is 12, then what is the original number?
(A) 28
(B) 39
(C) 82
(D) 89
(E) 93
(A) 28
(B) 39
(C) 82
(D) 89
(E) 93
04 Aug 2021, 02:59
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Explanation:
Sum of digit is 12
The only options which are possible 39, 93
option A,C,D are out
as we know, the original number is greater by 54.
IMO-E
Sum of digit is 12
The only options which are possible 39, 93
option A,C,D are out
as we know, the original number is greater by 54.
IMO-E
04 Aug 2021, 04:07
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Solution:
Solution:
There is a methodical way and hit and trial method to solve this question.
Methodical Way:
Let us assume the \(2-\)digit number \(= 10x+y\). Where \(x\) and \(y\) are ten's and unit's digit respectively.
So. the number after interchanging the digit will be \(= 10y+x\)
We are given that difference between original and digit reversed number \(= 54\). Thus we can say:
\(10x+y-(10y+x)=54\)
\(⇒ 10x+y-10y-x=54\)
\(⇒ 9x-9y=54\)
\(⇒ 9(x-y)=54\)
\(⇒ x-y=6\).......\((i)\). This is our equation \(1\).
We are also given that sum of digits \(= 12\). Thus we can say:
\(⇒ x+y=12\).....\((ii)\). This is our equation \(2\).
On adding the 2 equations we get: \(x-y+x-y=6+12\)
\(⇒ 2x=18\)
\(⇒ x=\frac{18}{2}\)
\(⇒ x=9\)
Plugging \(x=9\) in equation 2 we get,
\(⇒ x+y=12\)
\(⇒ 9+y=12\)
\(⇒ y=3\)
Hence the 2-digit number \(= 93\)
Hit and Trial Method:
Since sum of digits \(= 12\), answer can be either Option B or E
Looking at the second scenario, the difference between original and digit reversed number \(= 54\). We see that \(93-39=54\)
Hence the right answer is Option E.
Solution:
There is a methodical way and hit and trial method to solve this question.
Methodical Way:
Let us assume the \(2-\)digit number \(= 10x+y\). Where \(x\) and \(y\) are ten's and unit's digit respectively.
So. the number after interchanging the digit will be \(= 10y+x\)
We are given that difference between original and digit reversed number \(= 54\). Thus we can say:
\(10x+y-(10y+x)=54\)
\(⇒ 10x+y-10y-x=54\)
\(⇒ 9x-9y=54\)
\(⇒ 9(x-y)=54\)
\(⇒ x-y=6\).......\((i)\). This is our equation \(1\).
We are also given that sum of digits \(= 12\). Thus we can say:
\(⇒ x+y=12\).....\((ii)\). This is our equation \(2\).
On adding the 2 equations we get: \(x-y+x-y=6+12\)
\(⇒ 2x=18\)
\(⇒ x=\frac{18}{2}\)
\(⇒ x=9\)
Plugging \(x=9\) in equation 2 we get,
\(⇒ x+y=12\)
\(⇒ 9+y=12\)
\(⇒ y=3\)
Hence the 2-digit number \(= 93\)
Hit and Trial Method:
Since sum of digits \(= 12\), answer can be either Option B or E
Looking at the second scenario, the difference between original and digit reversed number \(= 54\). We see that \(93-39=54\)
Hence the right answer is Option E.
